Problem: Given the function $y=x^2+10x+21$, what is the least possible value of $y$?
Answer: When graphed, this function is a parabola that opens upwards. Thus, the minimum possible value of y occurs at the vertex of the parabola. The $x$ coordinate of the vertex is $\frac{-b}{2a}$. Substituting the given values, this yields $\frac{-10}{2}=-5$. Substituting this for $x$ gives the minimum value of $y$ to be \begin{align*}
y&=x^2+10x+21 \\
&=(-5)^2+10(-5)+21 \\
&=25+(-50)+21 \\
&=25-50+21 \\
&=-25+21 \\
&=\boxed{-4}
\end{align*}